Consider the following chemical reaction.

$$\ce{A + B -> Products}$$

If this is a zero-order reaction with respect to $\ce{A}$ and $\ce{B}$, then is there a half-time for this reaction? Or must we calculate the half-time for species $\ce{A}$ and $\ce{B}$ separately?

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    $\begingroup$ The concept of half life is only useful when the reaction is first order with respect to A and does not depend on anything else. In all other cases, though still technically applicable, it does not do any good. $\endgroup$ Commented Jul 30, 2023 at 15:45

1 Answer 1


For these reaction order you have

$$ -\frac{d|\ce{A}|}{dt} = -\frac{d|\ce{B}|}{dt} = k|\ce{A}|^0|\ce{B}|^0 = k$$

and the integrated velocity equations have the form

$$ |\ce{A}|_t = |\ce{A}|_0 - kt$$ $$ |\ce{B}|_t = |\ce{B}|_0 - kt$$

Using the definition of $t_{1/2}$ you will have

$$t_{1/2}(\ce{A}) = \frac{|\ce{A}|_0}{2k} $$ $$t_{1/2}(\ce{B}) = \frac{|\ce{B}|_0}{2k} $$

so, as you suggested, unless $|\ce{A}|_0 = |\ce{B}|_0$, the value of $t_{1/2}$ depends on which reactant you are using to estimate its value.

  • $\begingroup$ So there is no term like half time of reaction for this case right,? $\endgroup$
    – O Ji
    Commented Jul 30, 2023 at 16:18
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    $\begingroup$ @OJi There is mathematically infinite number of possible formal half life values, depending on initial concentrations of A and B. Such a parameter is useless but for very particular reaction settings. $\endgroup$
    – Poutnik
    Commented Jul 30, 2023 at 16:45
  • $\begingroup$ No I meant that for overall reaction Is there half life of reaction? Like if initially concentration of A is 2M and B is 4M, then is there some thing like half life of reaction? $\endgroup$
    – O Ji
    Commented Jul 31, 2023 at 3:10
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    $\begingroup$ @Oji I agree with Poutnik, In the case you mention where both concentrations are different I do not think there is a definition for an overall lifetime of the reaction. If the initial concentrations of the reactants differ, It always would depend on the concentration of one of the reactants. It just will create confusion. $\endgroup$
    – PAEP
    Commented Jul 31, 2023 at 10:44
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    $\begingroup$ Since one of the reactants will be exhausted first, one could define $$t_{1/2} := \frac{\min\left\{[\ce{A}]_0, [\ce{B}]_0\right\}}{2k} $$ $\endgroup$ Commented Jul 31, 2023 at 11:09

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